As in Figure 13, AB, CD intersect at point O, AD = AO, CB = CO, E, F, G are the midpoints of OD, OB, AC, respectively, connect GE, GF, EF, try to investigate △GEF

Because CD is the diameter of circle O, so DE ⊥ AC,

And G is the midpoint of AD, by the nature of the right triangle have:

EG=GD=1/2AD;

And OE=OD=radius

OG=OG

Therefore, △ OEG and △ ODG are congruent<

Therefore ∠OEG=∠ODG=90°

That is, OE⊥EG

Thereby we know that GE is the tangent to the circle O